Select the correct statement(s) about the compound `NO[BF_(4)]:`

Options

A. It has `5sigma and 2pi` bond
B. Nitrogen-oxygen bond length is higher than nitric oxide `(NO)`
C. It is a diamagnetic species
D. `B-F` bond length in this compound is lower than in `BF_(3)`

Explanation

To solve the question regarding the compound \( NO[BF_4] \), we will analyze the statements provided and determine their correctness based on the molecular structure and bonding characteristics of the compound.

### Step-by-Step Solution:

1. **Identify the Structure of \( NO[BF_4] \)**:
- The compound can be represented as \( NO^+ \) (nitric oxide cation) and \( BF_4^- \) (tetrafluoroborate anion).
- The structure consists of a nitrogen-oxygen bond in the cation and a boron atom surrounded by four fluorine atoms in the anion.

2. **Calculate the Bond Order of \( NO^+ \)**:
- The Lewis structure of \( NO^+ \) shows a nitrogen-oxygen bond with one sigma bond and two pi bonds.
- The bond order is calculated as follows:
\[ \text{Bond Order} = \frac{\text{Total number of bonds}}{\text{Number of sigma bonds}} = \frac{3}{1} = 3 \]

3. **Calculate the Bond Order of \( NO \)**:
- For \( NO \), the bond order is found to be 2.5, indicating that the bond length in \( NO \) is greater than that in \( NO^+ \).

4. **Analyze the Sigma and Pi Bonds in \( NO[BF_4] \)**:
- In \( NO[BF_4] \), we have:
- From \( NO^+ \): 1 sigma bond.
- From \( BF_4^- \): 4 sigma bonds (one for each \( B-F \) bond).
- Total sigma bonds = 1 (from \( NO^+ \)) + 4 (from \( BF_4^- \)) = 5 sigma bonds.
- Total pi bonds = 2 (from \( NO^+ \)).
- Therefore, the total bond count is 5 sigma and 2 pi bonds.

5. **Evaluate Each Statement**:
- **Statement A**: "It has five sigma and two pi bonds."
- **Correct**: As calculated, there are indeed 5 sigma and 2 pi bonds.
- **Statement B**: "Nitrogen-oxygen bond length is higher than that in nitric oxide."
- **Incorrect**: Since the bond order of \( NO^+ \) is greater than that of \( NO \), the bond length of \( NO^+ \) is less than that of \( NO \).
- **Statement C**: "It is a diamagnetic species."
- **Correct**: Both \( NO^+ \) and \( BF_4^- \) are diamagnetic, making the entire compound diamagnetic.
- **Statement D**: "BF bond length in this compound is lower than that in \( BF_3 \)."
- **Incorrect**: The presence of back-bonding in \( BF_3 \) results in shorter \( B-F \) bond lengths compared to those in \( BF_4^- \).

### Conclusion:
The correct statements about the compound \( NO[BF_4] \) are **A** and **C**.